Solve for x: (16/81)^(sin^2 x) + (8/9)^(cos^2 x) = 26/27
If It is $\displaystyle ( \frac{4}{9})^{\cos^2{x}} $ not
$\displaystyle ( \frac{8}{9})^{\cos^2{x}} $
Sub $\displaystyle (\frac{4}{9})^{\sin^2{x}} = u $then
$\displaystyle
u^2 + (4/9) u^{-1} = M
$
$\displaystyle u^3 - M u + (4/9) = 0 $
It becomes a cubic equation
after solving u $\displaystyle u = (\frac{4}{9})^{\sin^2{x}} $
$\displaystyle x = \arcsin{ \sqrt{ \frac{ \ln{u} }{ \ln(4/9) } } }$
My comment: it is not true that this is the plot of $\displaystyle y = (16/81)^{\sin^2 x} + (8/9)^{\cos^2 x}$. Rather, this is the plot of $\displaystyle y = (16/81)^{sin^2 x} + (8/9)^{cos^2 x}-26/27$.
Also: This function assumes a minimum value of $\displaystyle \tfrac{19}{18}\approx 0.235$ at $\displaystyle x=\tfrac{\pi}{2}+n\cdot\pi, n\in\mathbb{Z}$. But it is always $\displaystyle >0$.